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This problem is from a past qualifying exam.

Definition A group $G$ has exponent $e$ if $g^e=1$ for all $g\in G$.

Problem

Let $G=\left(\mathbb{Z}/n\mathbb{Z}\right)^\times$. Find all the integers $n$ for which $G$ has exponent $2$.

My work so far: Need to find $n$ such that $\overline{b}^2=\overline 1$ for $\overline{b} \in G$. So I need to find $n$ such that $n \mid (b^2-1)$ and $(b,n)=1$. Well, how do I now go about finding all the $n$'s. I tried a few examples like $n=1$ etc. But then I still have to deal with $b$. This kind of problem is new to me. Can somebody kindly show me what I've done wrong/right here?

May be suggest a way to go about finding these $n$'s.

Thank for your help.

Cousin
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    If $n = \prod_{k=1}^n p_k^{m_k}$ is the prime factorisation of $n$, then by the Chinese remainder theorem, you have $\mathbb{Z}/(n) \cong \prod_{k=1}^n \mathbb{Z}/(p_k^{m_k})$. So you need only consider prime powers. – Daniel Fischer Aug 08 '13 at 17:33
  • It looks to me that if a group $G$ has exponent $2$, then every non-identity element of $G$ has order 2, i.e. $G \cong \mathbb{Z}{2} \times \mathbb{Z}{2} \times \cdots \times \mathbb{Z}_{2}$. This should narrow it down a lot, I think. – Alex Wertheim Aug 08 '13 at 17:34

1 Answers1

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A few helpful facts, that can be learned here:

  1. $(\mathbb{Z}/n\mathbb{Z})^{\times}=\prod(\mathbb{Z}/p_i^{k_i}\mathbb{Z})^{\times}$, where $n=\prod p_i^{k_i}$ is the prime factorization of $n$.
  2. $(\mathbb{Z}/2^k\mathbb{Z})^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{k-2}\mathbb{Z}$ for $k\ge 2$.
  3. $(\mathbb{Z}/p^k\mathbb{Z})^{\times}\cong\mathbb{Z}/p^{k-1}(p-1)\mathbb{Z}$ for odd primes $p$.

I believe this should be enough to answer your question. I got values $n=3,4,6,8,12,$ and $24$, and nothing else.

If you don't want to rely on the complete strength of 2. and 3. above, you can reason as follows. First, the element $3$ will have order greater than $2$ in the group $(\mathbb{Z}/2^k\mathbb{Z})^{\times}$ for $k\ge 4$, from which it follows that only $k=2$ and $k=3$ give groups with exponent $2$. Also, the element $2$ will have order greater than $2$ in the group $(\mathbb{Z}/p^k\mathbb{Z})^{\times}$ for $p\ge 5$ and for $p=3$, $k\ge 2$, from which it follows that only $p=3$ and $k=1$ gives a group with exponent $2$. Now put these results together with 1. to obtain the values of $n$ listed above.

Jared
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